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In article <1992Aug17.141017.373@csc.canterbury.ac.nz> wft@math.canterbury.ac.nz (Bill Taylor) writes:
>I recall once seeing an article by Kemeny, in which he looked at a nonstandard
>model of the integers, in which there would be two (nonstandard) integers,
>n and n+2, both of which would be divisible by all standard integers. The
>hope was to be able to use this model to decide the twin-prime (or related)
>conjectures.
>
If the integer n is divisible by 4, then n + 2 is not. This fact
[obviously true in the standard model] can be proved in PA, and so
holds in non-standard models as well. 

It's a reasonable question (to which I don't know the answer) what the
result of Kemeny is that wft is misremembering.
>
>Thanks,      Bill Taylor.         wft@math.canterbury.ac.nz

Bob Solovay solovay@math.berkeley.edu

>>model of the integers, in which there would be two (nonstandard) integers,
>>n and n+2, both of which would be divisible by all standard integers. 

>It's a reasonable question (to which I don't know the answer) what the
>result of Kemeny is that wft is misremembering.

Sorreeee; bit of a boo-boo here. Brain damage my only excuse.

Should be:  there is a nonstandard integer n, which is divisible by all
standard integers. Thus  (n+1) and (n-1) are not divisible by any of them,
and are thus "standardly" twin primes. Of course, they may still be
composites in the whole nonstandard system. So this doesn't really help out
on the twin prime conjecture.

However; as I recall, Kemeny observed that (n-1), (n+1) are the *only* possible
primes in the "row" of numbers   .... , n-2 , n-1 , n , n+1 , n+2 , n+3 , ... 
Hence there are no 3-tuples of primes of type (p, p+2 , p+4) in the nonstandard
part of the model, (or any n-tuples with n>2). 

Thus, the n-tuple-primes conjectures of Hardy would all be false. The only
problem with this whole operation is showing that there can be a model with
*all* the "rows" having such an `n' in them. Kemeny reluctantly left this 
loop-hole unplugged.

Has anyone heard of the paper ? Or more precisely, followups to it ?

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            Bill Taylor              wft@math.canterbury.ac.nz 
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 Stop press ! I have just scaveged up a copy of Kemeny's paper.

It is "Undecidable problems of elementary number theory",
       Math Annalen, Bd. 135, (1958), pp 160-169.

I would still appreciate hearing of any follow-ups, though !

In article <1992Aug18.133344.385@csc.canterbury.ac.nz> wft@math.canterbury.ac.nz (Bill Taylor) writes:

>
>Thus, the n-tuple-primes conjectures of Hardy would all be false. The only
>problem with this whole operation is showing that there can be a model with
>*all* the "rows" having such an `n' in them. Kemeny reluctantly left this 
>loop-hole unplugged.
>

>--------------------------------------------------------------------------
>            Bill Taylor              wft@math.canterbury.ac.nz 
>--------------------------------------------------------------------------

	Kemeny's conjecture is false: If M is a non-standard model of
PA, then there is a non-standard integer n such that no integer in the
"standard row" of n is divisible by all standard integers.

	To see this, take n which is odd but is divisible by all
standard odd integers. If m differs from n by a standard integer, then
there must be a standard odd prime which does not divide the
difference. Hence m is not divisible by that prime.

>	Kemeny's conjecture is false: If M is a non-standard model of
>PA, then there is a non-standard integer n such that no integer in the
>"standard row" of n is divisible by all standard integers.
>
>	To see this, take n which is odd but is divisible by all
>standard odd integers. If m differs from n by a standard integer, then
>there must be a standard odd prime which does not divide the
>difference. Hence m is not divisible by that prime.

Looking at this bit...

>take n which is odd but is divisible by all standard odd integers.

...at first it wasn't clear to me that every nonstandard model would have
such an  n .  But I suppose this can be seen by taking any odd nonstandard k,
and forming  n = k(k-2)(k-4)(k-6).....  .

Presumably this operation can be made legitimate in some simple way ?
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            Bill Taylor              wft@math.canterbury.ac.nz 
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                  Kleeneness is next to Godelness.
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